Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

More documents

Recommendations

Info

88 ADVANCED ABSTRACT ALGEBRA rt + M = (r+M) (t+M) ∈ A* (because A* is an ideal of R/M) . rt ∈ A. R being commutative tr also belongs to A. Hence A is an ideal of R. If a ∈ M, then a+M = M ∈ R/M (since M is the zero element of R/M) a+M ∈ A* (since (1+M) (a+m) ∈ A* , a ∈ A A* being ideal of R/M) Therefore M ⊂ A ⊂ R . Let us suppose that A* ≠ {0} then there exists an element r+M of A* such that r +M ≠ M But r+M ∈ A* r ∈ A , r+M ≠ M r ∉ M A ≠ M. Thus we have proved that if A* ≠ {0}, then M ⊂ A ⊂ R Since M is maximal therefore, A = R. If r ∈ R then r ∈ A which implies that r +M ∈ A*. It follows therefore, that A* = R/M. We have proved therefore, that R/M has only two ideals {0} and R/M and hence R/M is a field. Conversely, let R/M is a field. Then R/M has only two ideals {0} and R/M itself. Hence A* = {0} or A* = R/M. If A* = {0} then A* = M Therefore, A = {r | r +M ∈ A*} = {r | r+M = M} = {r | r ∈ M} = M If A* = R/M then A = {r | r +M ∈ R/M} (Θ M is zero element of R/M) = {r | r ∈ M} = R . Therefore, R has only two ideals M and R. Hence M is a maximal ideal. Imbedding of a ring and an integral domain. Definition. If a ring R is isomorphic to a subring T of a ring S then R is called imbedded in S. The ring S is called extension or over ring of R. Theorem. Every ring R can be imbedded in a ring S with unit element. Proof. Let S be a set defined by S = Z × R = {(m,a) | m ∈ Z, a ∈ R}.
UNIT-IV 89 We define addition and multiplication in S as follow : (m, a) + (n, b) = (m+n, a+b) (m, a) (n, b) = (mn, na+ mb + ab) We now prove that S is a ring with unity under these binary operations. Let (m, a), (n, b), (p, c) ∈ S. Then (i) [(m,a) + (n, b)] + (p, c) = (m+n, a+b) + (p,c) = (m+n+p, a+b+c) = (m+(n+p), a+(b+c)) (by Associativity of R and Z) = (m,a) + (n+p, b+c) = (ma) + [(n,b) + (p,c)] (ii) (0,0) + (m,a) = (m,a) (m,a) + (0,0) = (m,a) Therefore (0,0) is additive, identity. (iii) Therefore (iv) (v) and Hence (vi) and (m,a) + (−m, −a) = (0,0) (−m,−a) + (m,a) = (0,0) (−m, −a) is the inverse of (m,a). (m,a) + (n,b) = (m+n, a+b) = (n+m, b+a) (by commutativity of R and Z) = (n,b) + (m,a) [(m,a) (n,b)] (p,c) = [mn, na + mb + ab] (p,c) = [(mn)p, p(na+ mb + ab) + mnc + c(na+mb+ab)] = [(mn)p, p(na) + p(mb) + p(ab) +(mn)c +(na)c+(mb)c+(ab)c] (m,a) [(n,b) (p,c)] = (m,a) [np, pb + nc + bc] = [m(np), anp + m(pb) + m(nc) + m(bc) +a(pb+nc+bc)] = [(mn)p, p(na) + p(mb) + p(ab) + (mn) c + (na) c + (mb)c + (ab) c] (by Associativity and commutativity of R and Z). (m,a) [(n,b) (p,c)] [(m,a) + (n,b)] (p,c) = [(m,a) (n,b)] (p,c) = (m+n, a+b)(p,c) = [(m+n)p, p(a+b) + (m+n)c + (a+b)c] = (mp+np, pa + pb + mc+ nc + ac + bc)
Page 1 and 2:
1 Advanced Abstract Algebra M.A./M.
Page 3 and 4:
Contents 3 Unit I 5 Unit II 49 Unit
Page 5 and 6: UNIT-I 5 Unit-I Definition Group A
Page 7 and 8: UNIT-I 7 2. In example 1, a 1 1 0 1
Page 9 and 10: UNIT-I 9 Solutions: 1. CGbg≠ a φ
Page 11 and 12: UNIT-I 11 ∴ Ha ⊆ H. To show let
Page 13 and 14: UNIT-I 13 Corollary 4: (Feremat's L
Page 15 and 16: UNIT-I 15 Definition: Group Homomor
Page 17 and 18: UNIT-I 17 Proof: Consider the diagr
Page 19 and 20: UNIT-I 19 ( G ) N G ( H ) ≅ H N D
Page 21 and 22: UNIT-1 21 Theorem 8. Let H be a sub
Page 23 and 24: UNIT-1 23 From (i) and (ii), we hav
Page 25 and 26: UNIT-I 25 1 Note that αβ = 2 2 1
Page 27 and 28: UNIT-1 27 In this case the length o
Page 29 and 30: UNIT-1 29 | n Theorem 18. The set o
Page 31 and 32: UNIT-I b = φ ( φ φ ) ( ) g h t x
Page 33 and 34: UNIT-I 33 This is impossible. So no
Page 35 and 36: UNIT-I 35 integers x and y. Note th
Page 37 and 38: UNIT-I 37 O(7) = 4, as 7 1 = 7, 7 2
Page 39 and 40: UNIT-I 39 Put a x mt , = b = x Then
Page 41 and 42: UNIT-I n iii. N = ∈Gs, show that
Page 43 and 44: UNIT-I 43 abab------ab = a n b n
Page 45 and 46: UNIT-I Q.20. Q.21. Q.22. Q.23. Q.24
Page 47 and 48: UNIT-I 47 Remark : If G is abelian,
Page 49 and 50: UNIT-IV 81 Definition. Let S be a s
Page 51 and 52: UNIT-IV 83 (r+s) − (r 1 +s 1 ) =
Page 53 and 54: UNIT-IV 85 and = ψ(r+K) + ψ(s+K)
Page 55: UNIT-IV 87 M ⊂ M′ ⊂ R M′ =
Page 59 and 60: UNIT-IV 91 be the set of the ordere
Page 61 and 62: UNIT-IV 93 (iii) a + b − a b
Page 63 and 64: UNIT-IV 95 and Z/A ~ Im(f) = f (z)
Page 65 and 66: UNIT-IV 97 If a n = 0 for all n ≥
Page 67 and 68: UNIT-IV 99 x 3 = (0, 0, 0, 1,…) .
Page 69 and 70: UNIT-IV 101 Then either (i) or n i
Page 71 and 72: UNIT-IV 103 deg f(x) = 0 and deg g(
Page 73 and 74: UNIT-IV 105 If x, y ∈ A , then th
Page 75 and 76: UNIT-IV 107 The possibility φ(r) <
Page 77 and 78: UNIT-IV 109 Lemma 4. If f(x) is an
Page 79 and 80: UNIT-IV 111 Proof. Let, if possible
Page 81 and 82: UNIT-II 49 Unit-II Definition: Comp
Page 83 and 84: UNIT-II 51 If r = 1, then G is simp
Page 85 and 86: UNIT-II 53 group of the Commutator
Page 87 and 88: UNIT-II di + 1 idi + 1 i di + 1 idi
Page 89 and 90: UNIT-II 57 we write the refinement
Page 91 and 92: UNIT-II 59 of length e in which eac
Page 93 and 94: UNIT-II 61 of nilpotency of G) , so
Page 95 and 96: UNIT-III 63 n.v = R i.e. abelian gr
Page 97 and 98: UNIT-III 65 Theorem 1 Let M, N be R
Page 99 and 100: UNIT-III 67 F ( BA ad 1'= = , 1a d(
Page 101 and 102: UNIT-III 69 Vector space over F. Le
Page 103 and 104: UNIT-III 71 1. J = only one linearl
Page 105 and 106: UNIT-III Hence J = −L NM 4 0 0 0
Page 107 and 108:
UNIT-III 75 Example 5 The direct su
Page 109 and 110:
UNIT-IV 77 A. B ≠ B. A. If IR, th
Page 111 and 112:
UNIT-IV 79 Example: (D, +, .) is a
Page 113 and 114:
UNIT-IV 81 Examples 1. The polynomi
Page 115 and 116:
UNIT-V 129 (Characteristic of a rin
Page 117 and 118:
UNIT-V 131 As F = p dΘ F ≅ Z p i
Page 119 and 120:
UNIT-V 133 In this case is called t
Page 121 and 122:
+ c c+ 0 0 1 c c+ 1 1 1 0 1+ c c c
Page 123 and 124:
UNIT-V 137 element of defines a per
show all

Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?